8.4 Comparison Tests

OBJECTIVES

When you finish this section, you should be able to:

  1. Use Comparison Tests for Convergence and Divergence (p. 576)
  2. Use the Limit Comparison Test (p. 577)

We can determine whether a series converges or diverges by comparing it to a series whose behavior we already know. Suppose \(\sum\limits_{k=1}^{\infty}b_{k}\) is a series of positive terms that is known to converge to \(B\), and we want to determine if the series of positive terms \(\sum\limits_{k=1}^{\infty }a_{k}\) converges. If term-by-term \(a_{k}\leq b_{k}\), that is, if \[ a_{1}\leq b_{1},~\ a_{2}\leq b_{2},\ldots , a_{n}\leq b_{n}\,, \ldots \]

then it follows that \[ \left( n\hbox{th} \hbox{ partial sum of }\sum\limits_{k=1}^{\infty }a_{k}\right) \leq \left( n\hbox{th} \hbox{ partial sum of }\sum\limits_{k=1}^{\infty }b_{k}\right) <B \]

This shows that the sequence of partial sums \(\{S_{n}\}\) of \(\sum\limits_{k=1}^{\infty }a_{k}\) is bounded, so by the General Convergence Test, \(\sum\limits_{k=1}^{\infty }a_{k}\) must also converge. This proves the Comparison Test for Convergence.

RECALL

The General Convergence Test: An infinite series of positive terms converges if and only if its sequence of partial sums is bounded.

576

THEOREM Comparison Test for Convergence

If \(0<a_{k}\leq b_{k}\) for all \(k\) and \(\sum\limits_{k=1}^{\infty}b_{k}\) converges, then \(\sum\limits_{k=1}^{\infty}a_{k}\) converges.

1 Use Comparison Tests for Convergence and Divergence

It is important to remember that the early terms in a series have no effect on the convergence or divergence of a series. In fact, the Comparison Test for Convergence is true if \(0<a_{n}\leq b_{n}\) for all \(n\geq N\), where \(N\) is some suitably selected integer. We use this in the next example by ignoring the first term.

Using the Comparison Test for Convergence

Show that the series below converges: \[ \sum\limits_{k=1}^{\infty }\frac{1}{k^{k}}=1+\frac{1}{2^{2}}+\frac{1}{3^{3}} +\cdots +\frac{1}{n^{n}}+\cdots \]

Solution We know that the geometric series \(\sum\limits_{k=1}^{\infty }\dfrac{1}{2^{k}}\) converges since \(\vert r\vert =\dfrac{1}{2}<1\). Now, since \(\dfrac{1}{n^{n}}\leq \dfrac{1}{2^{n}}\) for all \(n\geq 2\), except for the first term, each term of the series \(\sum\limits_{k=1}^{\infty }\dfrac{1}{k^{k}}\) is less than or equal to the corresponding term of the convergent geometric series. So, by the Comparison Test for Convergence, the series \(\sum\limits_{k=1}^{\infty }\dfrac{1}{k^{k}}\) converges.

NOTE

Do you see why \(\dfrac{1}{n^{n}}\leq \dfrac{1}{2^{n}}\) when \(n \geq 2\)? \(n\geq 2\Rightarrow {n}^{n}\geq {2}^{n}\Rightarrow\) \(\dfrac{{ 1}}{{n}^{n}}\leq \dfrac{1}{{2}^{n}}\).

NOW WORK

Problem 5.

In Example 1, you may have asked, “How did you know that the given series should be compared to \(\sum\limits_{k=1}^{\infty }\dfrac{1}{2^{k}}\)?” The easy answer is to try a series that you know converges, such as a geometric series with \(\vert r\vert <1\) or a \(p\)-series with \(p>1\). The honest answer is that only practice and experience will guide you.

There is also a comparison test for divergence. Suppose \(\sum\limits_{k=1}^{\infty }c_{k}\) is a series of positive terms we know diverges, and \(\sum\limits_{k=1}^{\infty }a_{k}\) is the series of positive terms to be tested. If term-by-term \(a_{k}\geq c_{k}\), that is, if \[ a_{1}\geq c_{1}, a_{2}\geq c_{2},\ldots, a_{n}\geq c_{n},\ldots \]

then it follows that \[ \left( n \hbox{th} \hbox{ partial sum of }\sum\limits_{k=1}^{\infty}a_{k}\right) \geq \left(n\hbox{th} \hbox{ partial sum of }\sum\limits_{k=1}^{\infty }c_{k}\right) \]

But we know that the sequence of partial sums of \(\sum\limits_{k=1}^{\infty}c_{k}\) is unbounded. So, the sequence of partial sums of \(\sum\limits_{k=1}^{\infty }a_{k}\) is also unbounded, and \(\sum\limits_{k=1}^{\infty }a_{k}\) diverges.

THEOREM Comparison Test for Divergence

If \(0<c_{k}\leq a_{k}\) for all \(k\) and \(\sum\limits_{k=1}^{\infty}c_{k}\) diverges, then \(\sum\limits_{k=1}^{\infty }a_{k}\) diverges.

577

Using the Comparison Test for Divergence

Show that the series \(\sum\limits_{k=1}^{\infty }\dfrac{k+3}{k(k+2)}\) diverges.

Solution Since \(\dfrac{k+3}{k(k+2)}\) has a factor \(\dfrac{1}{k}\), we choose to compare the given series to the harmonic series \(\sum\limits_{k=1}^{\infty }\dfrac{1}{k}\), which diverges. \[ \frac{n+3}{n(n+2)}=\left( {\frac{n+3}{n+2}}\right) \left( {\frac{1}{n}} \right) >\frac{1}{n} \]

It follows from the Comparison Test for Divergence that \(\sum\limits_{k=1}^{\infty }\dfrac{k+3}{k(k+2) }\) diverges.

NOW WORK

Problem 9.

2 Use the Limit Comparison Test

The Comparison Tests for Convergence and Divergence are algebraic tests that require certain inequalities hold. Obtaining such inequalities can be challenging. The Limit Comparison Test requires certain conditions on the limit of a ratio. You may find this comparison test easier to use.

THEOREM Limit Comparison Test

Suppose \(\sum\limits_{k=1}^{\infty }\,a_{k}\) and \(\sum\limits_{k=1}^{\infty }\,b_{k}\) are both series of positive terms.

If \(\lim\limits_{n\,\rightarrow \,\infty }\,\dfrac{a_{n}}{b_{n}}=L\), \(0<\) \(L<\infty ,\) then both series converge or both series diverge.

The Limit Comparison Test gives no guidance if \(\lim\limits_{n\,\rightarrow \,\infty }\dfrac{{ a}_{n}}{{ b}_{n}}\) \(=0\) or \(\lim\limits_{n\,\rightarrow \,\infty }\dfrac{{ a}_{n}}{{ b}_{n}}=\infty\) or \(\lim\limits_{n\rightarrow \infty }\dfrac{a_{n}}{b_{n}}\) does not exist.

Proof

If \(\lim\limits_{n\,\rightarrow \,\infty }\,\dfrac{a_{n}}{b_{n}}=L\,{>}\,0\), then for any \(\varepsilon\,{>}\,0,\) there is a positive real number \(N\) so that \(\left\vert \dfrac{a_{n}}{b_{n}}-L\right\vert <\varepsilon ,\) for all \(n>N.\) If we choose \(\varepsilon =\dfrac{L}{2},\) then \[ \begin{eqnarray*} \begin{array}{rl@{\qquad\hspace{18.2pt}}l} \left\vert \dfrac{a_{n}}{b_{n}}-L\right\vert &< \dfrac{L}{2} & \hbox{for all }n>N \\[13pt] \dfrac{L}{2}<\dfrac{a_{n}}{b_{n}} &< \dfrac{3L}{2} & \hbox{for all }n>N\, \end{array} \end{eqnarray*} \]

Since \(b_{n}>0\), \[ \begin{array}{rl@{\qquad}l} \dfrac{L}{2}b_{n}<a_{n} & <\dfrac{3L}{2}b_{n} & \hbox{for all }n>N \end{array} \]

If \(\sum\limits_{k=1}^{\infty }a_{k}\) converges, by the Comparison Test for Convergence, the series \(\sum\limits_{k=1}^{\infty }\left(\dfrac{L}{2}b_{k}\right)\) \(=\dfrac{L}{2}\sum\limits_{k=1}^{\infty }b_{k}\) also converges. It follows that the series \(\sum\limits_{k=1}^{\infty }b_{k}\) is also convergent.

If \(\sum\limits_{k=1}^{\infty }b_{k}\) converges, then so does \(\sum\limits_{k=1}^{\infty }\left(\dfrac{3L}{2}b_{k}\right).\) Since \(a_{n}<\dfrac{3L}{2}b_{n},\) for all \(n>N,\) by the Comparison Test for Convergence, \(\sum\limits_{k=1}^{\infty }a_{k}\) also converges.

Therefore, \(\sum\limits_{k=1}^{\infty }a_{k}\) and \(\sum\limits_{k=1}^{\infty}b_{k}\) converge together. Since one cannot converge and the other diverge, they must diverge together.

RECALL

If \(c\) is a nonzero real number and if \(\sum\limits_{k\,=\,1}^{\infty}{a}_{k}={S}\) is a convergent series, then the series \(\sum\limits_{k\,=\,1}^{\infty} ({ca}_{k})\) also converges. Furthermore, \(\sum\limits_{k\,=\,1}^{\infty}({ca}_{k})={c}\sum\limits_{k\,=\,1}^{\infty }{a}_{k}={cS}\).

578

The Limit Comparison Test is quite versatile for comparing algebraically complex series to a \(p\)-series. To find the correct choice of the \(p\)-series to use in the comparison, examine the behavior of the terms of the series for large values of \(n.\) For example,

Using the Limit Comparison Test

Determine whether the series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{2k^{3/2}+5}\) converges or diverges.

Solution We choose an appropriate \(p\)-series to use for comparison by examining the behavior of the series for large values of \(n\): \[ \begin{eqnarray*} \dfrac{1}{2n^{3/2}+5}=\dfrac{1}{n^{3/2}\left( 2+\dfrac{5}{n^{3/2}}\right) }= \dfrac{1}{n^{3/2}}\left( \dfrac{1}{2+\dfrac{5}{n^{3/2}}}\right) & \underset{\underset{\color{#0066A7}{{\hbox{for large }\hbox{\(n\)}}}}{{\color{#0066A7}{{\uparrow }}}} }{\approx} & \dfrac{1}{n^{3/2}}\left(\dfrac{1}{2}\right)\\ \end{eqnarray*} \]

This leads us to choose the \(p\)-series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k^{3/2}},\) which converges, and use the Limit Comparison Test with \[ a_{n}=\dfrac{1}{2n^{3/2}+5} \qquad \hbox{and}\qquad b_{n}=\dfrac{1}{n^{3/2}} \] \[ \lim\limits_{n\rightarrow \infty }\dfrac{a_{n}}{b_{n}}=\lim_{n\,\rightarrow \,\infty }\frac{\dfrac{1}{2n^{3/2}+5}}{\dfrac{1}{n^{3/2}}} =\lim\limits_{n\rightarrow \infty }\frac{n^{3/2}}{2n^{3/2}+5} =\lim\limits_{n\rightarrow \infty }\dfrac{1}{2+\dfrac{5}{n^{3/2}}}=\frac{1}{2} \]

Since the limit is a positive number and the \(p\)-series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k^{3/2}}\) converges, then by the Limit Comparison Test, \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{2k^{3/2}+5}\) also converges.

NOW WORK

Problem 15.

579

Using the Limit Comparison Test

Determine whether the series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{3\sqrt{k}+2}{\sqrt{k^{3}+3k^{2}+1}}\) converges or diverges.

Solution We choose a \(p\)-series for comparison by examining how the terms of the series behave for large values of \(n\): \[ \begin{eqnarray*} \frac{3\sqrt{n}+2}{\sqrt{n^{3}+3n^{2}+1}}=\frac{\sqrt{n}\,\left( 3+\dfrac{2}{ \sqrt{n}}\right) }{\sqrt{n^{3}}\left( \sqrt{1+\dfrac{3}{n}+\dfrac{1}{n^{3}}} \right) }=\dfrac{n^{1/2}}{n^{3/2}}\frac{3+\dfrac{2}{\sqrt{n}}}{\sqrt{1+ \dfrac{3}{n}+\dfrac{1}{n^{3}}}} & \underset{\underset{\color{#0066A7}{{\hbox{for large }\hbox{\(n\)}}}}{{\color{#0066A7}{{\uparrow}}}}}{\approx} & \dfrac{1}{n}(3) \\ \end{eqnarray*} \]

So, we compare the series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{3\sqrt{k~}+2}{\sqrt{k^{3}+3k^{2}+1}}\) to the harmonic series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k}\), which diverges, and use the Limit Comparison Test with \[ a_{n}=\frac{3\sqrt{n}+2}{\sqrt{n^{3}+3n^{2}+1}}\qquad \hbox{ and }\qquad b_{n}=\dfrac{1}{n} \] \[ \begin{eqnarray*} \lim\limits_{n\rightarrow \infty }\dfrac{a_{n}}{b_{n}} &=& \lim_{n\,\rightarrow \,\infty }\frac{\dfrac{3\sqrt{n}+2}{\sqrt{n^{3}+3n^{2}+1}}}{\dfrac{1}{n}} =\lim\limits_{n\rightarrow \infty }\frac{n\left( 3\sqrt{n}+2\right) }{\sqrt{ n^{3}+3n^{2}+1}}=\lim\limits_{n\rightarrow \infty }\frac{3n^{3/2}+2n}{\sqrt{n^{3}+3n^{2}+1}}\\[7pt] &=&\lim\limits_{n\rightarrow \infty }\dfrac{3n^{3/2}}{n^{3/2}}=3 \end{eqnarray*} \]

Since the limit is a positive real number and the \(p\)-series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k}\) diverges, then by the Limit Comparison Test, \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{3\sqrt{k~}+2}{\sqrt{k^{3}+3k^{2}+1}}\) also diverges.

NOW WORK

Problem 17.

Using comparison tests to determine whether a series converges or diverges requires that you know convergent and divergent series to use in the comparison. Table 3 lists some series we have already encountered that are useful for this purpose.

Table 3: TABLE 3
Series Convergent Divergent
The geometric series \(\sum\limits_{k\,=\,1}^{\infty}ar^{k-1}\) \(\vert ~r \vert <1\) \(\vert ~r \vert \geq 1\)
The harmonic series \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{1}{k}\) Divergent
The \(p\)-series \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{1}{k^{p}}\) \(p>1\) \(0<p\leq 1\)
The series \(\sum\limits_{k\,=\,1}^{\infty}\dfrac{1}{k^{k}}\) Convergent